Tiny exoplanet is smaller than Mercury (and probably hotter, too)

Planet takes a mere 13 days to orbit its star.

Exoplanets in the Kepler-37 system, compared with Solar System bodies. The smallest of the planets, Kepler-37b, is only slightly larger than the Moon, and noticeably smaller than Mercury.

NASA/Ames/JPL-Caltech

What's the smallest exoplanet we can detect from Earth? Smaller bodies are inherently harder to observe than larger ones, but the orbiting Kepler telescope has enabled astronomers to identify a number of small planet candidates, including a few smaller than Earth, by measuring the fluctuations in the host star's light as the planets pass in front of it.

Now researchers may have found the smallest exoplanet yet, a world with a diameter about 80 percent of Mercury's. This planet candidate, named Kepler-37b, orbits very close to its star: its orbital radius is about 1/4 the size of Mercury's, so it takes only about 13 days to zip around. Thomas Barclay and collaborators also identified two other planets in the same system—labeled Kepler-37c and Kepler-37d—one of which is slightly larger than Mars, and the other which has twice Earth's diameter. While these worlds are all in very small orbits, their existence contributes to our understanding of the variety of planets that can form.

The Kepler telescope works by collecting light from a huge number of stars in a single patch of the sky. By searching for periodic fluctuations in the light from an individual star, astronomers can find exoplanets as they transit, crossing in front of the disk of the star and blocking a tiny amount of its light. The rapidity of the decrease in light and the fraction of the star's light that is lost to the eclipse reveal the size of the planet, while time between transits indicates the duration of the orbit. Since these effects are small, researchers use sophisticated computer algorithms to pick out exoplanet signals from other fluctuations, such as starspots (sunspots on other stars) .

Because transit observations don't rely on exoplanet mass, they allow astronomers to locate smaller worlds. A small orbit means a slightly larger eclipse, whatever the size of the planet. That's why the Earth-sized exoplanets detected so far are typically much closer to their star than Earth is.

The Kepler-37 system fits this pattern. Its three planets—denoted Kepler-37b, c, and d in order of their distance from the host star—are all much closer to the star than Mercury. The star itself, labeled Kepler-37a, is smaller, cooler, and therefore fainter than the Sun, which also assisted in detecting its planets: less brightness means the exoplanet transits show up more starkly.

Kepler-37d was the easiest to spot, having a diameter approximately twice that of Earth's and an orbit roughly half of Mercury's. (Those factors place the world in the "hot superearth" category, as the planet likely has a rocky surface.) Moving closer to the star, the astronomers measured Kepler-37c's diameter as about 75 percent of Earth's, placing it between Mars and Venus in size. They found its orbit was about 1/3 of Mercury's.

The innermost and smallest planet, Kepler-37b, also had the most ambiguous data. (Frankly, when I looked at their data, I had to take their word for it that there was actually a transit signal hiding in there, but that's what computer algorithms are for.) This world has a diameter only about 30 percent of Earth's, making it measurably smaller than Mercury's 38 percent. Its orbit is blisteringly small—10 percent of Earth's orbit, or 26 percent of Mercury's orbit—and it zips around its star in a mere 13 days.

With its small size and proximity to Kepler-37a, this planet is probably rocky and devoid of surface water or atmosphere. It also may be tidally locked, presenting the same face to its star, just as the same side of the Moon always faces Earth.

The probable discovery of a sub-Mercury-sized exoplanet opens up the possibility of finding more, and giving us a clearer picture of the diversity of planetary bodies. Kepler-37b wasn't a surprise: planet formation models predict the existence of such tiny bodies. However, locating them helps astronomers paint more details into our picture of planetary systems, which are far more varied and wonderful than could be guessed from our Solar System alone.

39 Reader Comments

A small orbit means a slightly larger eclipse, whatever the size of the planet

I'm a bit confused by this statement from the article - while technically this is correct (due to perspective, objects closer to the observer appear larger than objects further away) in this case the observer (Kepler) is essentially so massively far away from the star+planet system that there would be no observable size difference between a planet transiting 10 million miles away from its star and a planet 100 million miles away.

There would obviously be a timing difference (the planet further away will take longer to transit), but the occulting disk would essentially be the same size in both cases, so the dip in brightness will also be the same.

A small orbit means a slightly larger eclipse, whatever the size of the planet

I'm a bit confused by this statement from the article - while technically this is correct (due to perspective, objects closer to the observer appear larger than objects further away) in this case the observer (Kepler) is essentially so massively far away from the star+planet system that there would be no observable size difference between a planet transiting 10 million miles away from its star and a planet 100 million miles away.

There would obviously be a timing difference (the planet further away will take longer to transit), but the occulting disk would essentially be the same size in both cases, so the dip in brightness will also be the same.

What is meant is that assuming 2 planets of the same diameter, one being closer to the star than the other, the closer one blots out more of the star's visible "disk".

Its orbit being closer in will likely be much smaller and will indicate a much faster revolution around the star it orbits. Its transit across the disk of the star will be much quicker than the same sized plant further out from the star.

A small orbit means a slightly larger eclipse, whatever the size of the planet

I'm a bit confused by this statement from the article - while technically this is correct (due to perspective, objects closer to the observer appear larger than objects further away) in this case the observer (Kepler) is essentially so massively far away from the star+planet system that there would be no observable size difference between a planet transiting 10 million miles away from its star and a planet 100 million miles away.

There would obviously be a timing difference (the planet further away will take longer to transit), but the occulting disk would essentially be the same size in both cases, so the dip in brightness will also be the same.

Considering the distances involved to begin with, and the minuteness of light variations detected by Kepler, it's not hard to believe orbital distance could have a measurable impact (from Kepler's point of view) on the light blocked by the planet.

What I don't understand is why objects with a smaller orbit would have a larger eclipse, all else being equal. Given two planets of equal size and equal star brightness, wouldn't the one closer to Kepler (the one with the larger orbit, not smaller orbit) have a larger apparent eclipse? The quote you linked above from the article seems to indicate the inverse is true.

That's like saying holding a pencil 6 inches in front of your eye and blocking the TV screen will let more light in to your eye than taping the pencil directly to the TV screen. This does not make sense to me.

A small orbit means a slightly larger eclipse, whatever the size of the planet

I'm a bit confused by this statement from the article - while technically this is correct (due to perspective, objects closer to the observer appear larger than objects further away) in this case the observer (Kepler) is essentially so massively far away from the star+planet system that there would be no observable size difference between a planet transiting 10 million miles away from its star and a planet 100 million miles away.

There would obviously be a timing difference (the planet further away will take longer to transit), but the occulting disk would essentially be the same size in both cases, so the dip in brightness will also be the same.

Considering the distances involved to begin with, and the minuteness of light variations detected by Kepler, it's not hard to believe orbital distance could have a measurable impact (from Kepler's point of view) on the light blocked by the planet.

What I don't understand is why objects with a smaller orbit would have a larger eclipse, all else being equal. Given two planets of equal size and equal star brightness, wouldn't the one closer to Kepler have a larger apparent eclipse? The quote you linked above from the article seems to indicate the inverse is true.

That's like saying holding a pencil 6 inches in front of your eye and blocking the TV screen will let more light in to your eye than taping the pencil directly to the TV screen. This does not make sense to me.

I don't understand your question. A smaller orbit implies closer to its host star, consistent with the article and your belief.

I don't understand your question. A smaller orbit implies closer to its host star, consistent with the article and your belief.

The article states:

Quote:

A small orbit means a slightly larger eclipse, whatever the size of the planet

I'm trying to understand how that is possible. As my example stated above, if you have two planets equal in size orbiting stars of equal light magnitude and size, wouldn't the planet orbiting further away from its star (i.e. closer to the observer, Kepler, in this case) appear to have a larger eclipse compared to the planet that orbits more closely to its star?

I used the pencil/TV example to illustrate my point, but here's a better one, maybe:

When a solar eclipse occurs, the Moon blocks the light from the Sun as viewed from Earth. In this case, the Moon isn't orbiting the sun, but once they're aligned, that point is irrelevant. If the Moon orbited further away from the Earth, when a solar eclipse occurred, the Moon would appear further away from Earth, therefore smaller, and would block less light from the Sun, therefore causing less of the Sun to darken.

The quote from the article is indicating the opposite is true, and I'm not understanding how that's possible.

Is it the standard to name the planets in order of distance from their star instead of order discovered? I understand that people will assume a relationship between serially named planets and distance from their star, but what happens when you look at the system again with a better telescope or recheck the data with better algorithms and find another planet between b and c? Do you rename all the planets? If you don't, then it seems like you would no longer be able to rely on that assumption. If you do rename them, you confuse the hell out of people.

Is this something only someone from outside the field would even worry about?

What is meant is that assuming 2 planets of the same diameter, one being closer to the star than the other, the closer one blots out more of the star's visible "disk".

That's the bit I don't understand - like I said, this is technically true, but with the distances involved, this is not measurable.

Star is at 200 lightyears from us (and the Kepler observatory) - assuming two exactly identically sized planets, one at 10 million km from the star, the other at 100 million km.

1 lightyear is 10 million million km - the difference in distance between Kepler and both planets is 90 million kom - so the difference in apparent size between both planets will be 90 million divided by 2000 million million - roughly one part in about 2 million - we wouldn't even be able to measure a difference like that on planets in our own solar system, let alone 200 lightyears away.

A small orbit means a slightly larger eclipse, whatever the size of the planet

That's like saying holding a pencil 6 inches in front of your eye and blocking the TV screen will let more light in to your eye than taping the pencil directly to the TV screen. This does not make sense to me.

Think of it this way. Hold a flashlight. Put your hand in front of it. Do you cast a bigger shadow the closer or further away you pit your hand?

Pencil in front of eye is an invalid analogy because the amount of light entering your eye is orthogonal to the eclipse.

What is meant is that assuming 2 planets of the same diameter, one being closer to the star than the other, the closer one blots out more of the star's visible "disk".

That's the bit I don't understand - like I said, this is technically true, but with the distances involved, this is not measurable.

Star is at 200 lightyears from us (and the Kepler observatory) - assuming two exactly identically sized planets, one at 10 million km from the star, the other at 100 million km.

1 lightyear is 10 million million km - the difference in distance between Kepler and both planets is 90 million kom - so the difference in apparent size between both planets will be 90 million divided by 2000 million million - roughly one part in about 2 million - we wouldn't even be able to measure a difference like that on planets in our own solar system, let alone 200 lightyears away.

As I understand it, Kepler is that sensitive to light variations. While your logic is sound, you're making an assumption that Kepler can't detect such a small variation, but my understanding is that it can.

A small orbit means a slightly larger eclipse, whatever the size of the planet

That's like saying holding a pencil 6 inches in front of your eye and blocking the TV screen will let more light in to your eye than taping the pencil directly to the TV screen. This does not make sense to me.

Think of it this way. Hold a flashlight. Put your hand in front of it. Do you cast a bigger shadow the closer or further away you pit your hand?

Pencil in front of eye is an invalid analogy because the amount of light entering your eye is orthogonal to the eclipse.

Edit: After re-reading what you said, I'm taking back my original response. The size of the shadow cast by something blocking the light source behind it is irrelevant when you think about it. The shadow does not prevent the observer from seeing the light source leaking around the object at all, regardless of whether the shadow it is casting is bigger or smaller. The only thing the observer sees is the light passing around the object that is blocking some of the light, and the closer that object is to the observer (the larger it's orbit around the light source) the more light it will block from the point of view of the observer. I still think the article is wrong and/or needs to further explain the science behind the claim that objects closer to the light source block more light.

Original Response below for reference:

Spoiler: show

Thank you, that makes sense. However, I'm having difficulty resolving what you said, with my Solar eclipse analogy. For example, shouldn't your explanation mean that a transit of Venus across the Sun be more apparent to a Kepler-like observer than a transit of the Moon? Or am I confusing physical light blocking versus a shadow originating at a long distance, with Kepler detecting the latter, not the former?

A small orbit means a slightly larger eclipse, whatever the size of the planet

That's like saying holding a pencil 6 inches in front of your eye and blocking the TV screen will let more light in to your eye than taping the pencil directly to the TV screen. This does not make sense to me.

Think of it this way. Hold a flashlight. Put your hand in front of it. Do you cast a bigger shadow the closer or further away you pit your hand?

Pencil in front of eye is an invalid analogy because the amount of light entering your eye is orthogonal to the eclipse.

Thank you, that makes sense. However, I'm having difficulty resolving what you said, with my Solar eclipse analogy. For example, shouldn't your explanation mean that a transit of Venus across the Sun be more apparent to a Kepler-like observer than a transit of the Moon? Or am I confusing physical light blocking versus a shadow originating at a long distance, with Kepler detecting the latter, not the former?

My understanding (based on Physics for non science majors 10 years ago) is that an object will block less light the further it is from the source because light spreads out as it travels. The closer the object is to the source, the tighter the light beam, and the more light that is blocked.

I think from that distance the effect of the eclipse is probably largely the same.

Kepler is indeed very sensitive, in order to detect the light variation of a small planet transiting over a star, but that ratio is roughly the same as the size of the star to the size of the planet. Using sun and earth as an example, that'd be roughly 1:12000 (i.e. viewing from infinitely far away, the earth would obscure 1/12000 of the sun's light). However, if, say, someone using a Kepler observing the solar system from four light years away (~250000 AU), if the earth is now orbiting at 2 AU from the sun instead of 1 AU, it'd look only 1:250000 larger (EDIT: to put it more clearly, it's something like 250000 to 250001) to the observer. That's a much smaller ratio than the ratio of the cross section between the earth and the sun.

One thing that struck me about the article was the orbit time. 13 days is really short. If you had two objects of similar size orbiting in each others L3 point, you would have half the apparent orbit time. In this case, 2 objects orbiting at 26 days appearing to be one orbiting at 13 days. This is highly unlikely in any star system, but considering the unlikely things we've seen so far, something like this will probably pop up eventually.

>>> A small orbit means a slightly larger eclipse, whatever the size of the planet. That's why the Earth-sized exoplanets detected so far are typically much closer to their star than Earth is.

That's the wrong reason. Kepler detects planets with short orbital periods because it's been making observations for less than four years. For example, to detect a planet with a period like that of Saturn (29 years) would require at least 29 years of observations--and practically many, many more to confirm that the transits seen have a regular period.

Thank you, that makes sense. However, I'm having difficulty resolving what you said, with my Solar eclipse analogy. For example, shouldn't your explanation mean that a transit of Venus across the Sun be more apparent to a Kepler-like observer than a transit of the Moon? Or am I confusing physical light blocking versus a shadow originating at a long distance, with Kepler detecting the latter, not the former?

The star is far enough away to essentially be a pinpoint of light (unlike the moon/venus example, where the sun is a disk of light). When you have a pinpoint of light, you block more light when you're closer. To prove it to yourself, draw a few lines radiating out from a point, then two circles at two different distances. The closer circle touches (ie. blocks) more lines. I see your point, though, and admit that I'm no physicist or astronomer, so am not sure when one effect might take precedence over the other.

My understanding (based on Physics for non science majors 10 years ago) is that an object will block less light the further it is from the source because light spreads out as it travels. The closer the object is to the source, the tighter the light beam, and the more light that is blocked.

There are no light beams involved here. It doesn't matter that a closer planet blocks more light if that light would miss us anyway. In a solar eclipse, the moon blocks far less light than Venus does in a transit, but which produces the stronger signal?

The article has things backwards, a smaller orbital radius gives a (very slightly, at the distances involved) smaller eclipse, and it's quite wrong about this having anything to do with the detection rate favoring small orbits, which is purely down to them producing more transits to work with, reducing the effects of noise and variations in the star itself.

I've previously read that planets are detected by seeing "wobble" in the stars position due to gravity. Bigger wobble = bigger planet, this we can measure mass. Works for huge planets, not so well for small ones.

We can detect planets based on seeing a decrease in light output (partial eclipse). With a known size of a star, we can measure the % of light drop. I've previously read that this measurement relates to the size of the planet. If you can correlate with mass (by measuring the wobble), you can now also get density.

Now I read that the second measurement can be used to measure the size of the planets orbital radius. How can we tell the difference between orbital radius and planet size? I.e. how do we tell the difference between a small planet close to the star, and a large planet further away?

And, damn, I just answered it. For anyone else who is wondering, it's one of Kepler's laws (#2 I think). We can measure the transit duration and get a good idea of the distance.

The odds to see this wasn't high, but these types of finds are invaluable. The frequency of such small bodies, which we otherwise can't see (the star was bright, large and extra calm), must be staggering.

I don't understand your question. A smaller orbit implies closer to its host star, consistent with the article and your belief.

The article states:

Quote:

A small orbit means a slightly larger eclipse, whatever the size of the planet

I'm trying to understand how that is possible. As my example stated above, if you have two planets equal in size orbiting stars of equal light magnitude and size, wouldn't the planet orbiting further away from its star (i.e. closer to the observer, Kepler, in this case) appear to have a larger eclipse compared to the planet that orbits more closely to its star?

You don't have a diffuse, extended source where light converges on the observer from several directions due to the geometry. You have a point source that diverges from the source (as always).

Is it the standard to name the planets in order of distance from their star instead of order discovered? I understand that people will assume a relationship between serially named planets and distance from their star, but what happens when you look at the system again with a better telescope or recheck the data with better algorithms and find another planet between b and c?

They are named in order of detection, so only when there are several in a release round a specific star can the researchers use the "natural" order (and they don't have to). In other words, later planets are plunked in as they are discovered, no name change of old ones necessary.

I don't understand your question. A smaller orbit implies closer to its host star, consistent with the article and your belief.

The article states:

Quote:

A small orbit means a slightly larger eclipse, whatever the size of the planet

I'm trying to understand how that is possible. As my example stated above, if you have two planets equal in size orbiting stars of equal light magnitude and size, wouldn't the planet orbiting further away from its star (i.e. closer to the observer, Kepler, in this case) appear to have a larger eclipse compared to the planet that orbits more closely to its star?

You don't have a diffuse, extended source where light converges on the observer from several directions due to the geometry. You have a point source that diverges from the source (as always).

Or am I confusing physical light blocking versus a shadow originating at a long distance, with Kepler detecting the latter, not the former?

Yes.

I did read that article hoping to find an answer, and I appreciate your response, but I'm still not seeing a definitive, physical answer as to why "the smaller the orbit, the closer the planet to the star, the larger the blocking" is true. Can you elaborate?

You don't have a diffuse, extended source where light converges on the observer from several directions due to the geometry.

Um, no. This is exactly what you do have. That you can't resolve the extended source is irrelevant, you need only measure the total light received from it.

Torbjörn Larsson, OM wrote:

You have a point source that diverges from the source (as always).

So the smaller the orbit, the closer the planet to the star, the larger the blocking.

Your conclusion does not follow. A point source would actually blink in and out as the planet eclipsed it, not giving that U-shaped curve and not giving different results depending on orbital distance. You are confusing the amount of light hitting the planet with the amount of light it is occluding from our observation point.

While many of the comments said the article is wrong on this statement "A small orbit means a slightly larger eclipse," It got me a bit confused too in the beginning. But I got this light bulb in back of my head telling me that this article couldn't be wrong.

So I did a bit digging and thinking and reread the article. I noticed the article did mention the star itself is "smaller, cooler" than our Sun but without mentioning the exact size of it. So I figured the host star 37a may be not much larger than the 37d - the largest of the three planets.

If the host star Kepler-37a is not much larger than the Kepler-37d, then this statement "a small orbit means a slightly larger eclipse" may be a valid statement after all.

Say for an example: The size of our Sun equivalent to the size of a basketball, and our Earth to the size of a dime, and the moon to the size of a bean. And if 37a - equivalent to a quarter, 37b to a dime, 37c to a penny, and 37d equivalent to the size of a nickel, in this order. When you placed 37d - a nickel in front of 37a a quarter and from this great distance there to Earth, what this view from Kepler is that "its eclipse is slightly larger" is true, since the quarter is not much larger than the nickel, the nickel could blocked out more light from the quarter in that closer distance, than how a dime trying to block out the light from the size of a basketball is an impossible in this example. So it may give a whole different effect when viewed from an angle light years away.

Quote:

The star itself, labeled Kepler-37a, is smaller, cooler, and therefore fainter than the Sun, which also assisted in detecting its planets: less brightness means the exoplanet transits show up more starkly.

When you placed 37d - a nickel in front of 37a a quarter and from this great distance there to Earth, what this view from Kepler is that "its eclipse is slightly larger" is true, since the quarter is not much larger than the nickel, the nickel could blocked out more light from the quarter in that closer distance, than how a dime trying to block out the light from the size of a basketball is an impossible in this example.

You simply give your conclusion without reasoning that actually leads to it. And in fact, no, it *isn't* impossible for a dime to block out the light from an object the size of a basketball, solar eclipses are a demonstration of such a case.

Given an occluding body smaller than the emitting body and an observer at a fixed distance, there's always a distance beyond which the occluding body completely blocks the emitting body. At less than that distance, the occluding body will block some amount proportional to its angular area as seen from the observer's point of view, the amount blocked reaching a *minimum* at an orbital distance of zero.

Diffraction throws a minor wrench into this with the spot of Arago, but making it significant would likely require a planet large enough and aligned well enough to nearly perfectly eclipse its star, and it wouldn't lead to the described outcome anyway.

What is meant is that assuming 2 planets of the same diameter, one being closer to the star than the other, the closer one blots out more of the star's visible "disk".

Its orbit being closer in will likely be much smaller and will indicate a much faster revolution around the star it orbits. Its transit across the disk of the star will be much quicker than the same sized plant further out from the star.

Distance from the star does not change the amount of light that the planet blocks from our perspective. The other effect matters. Kepler is biased to find planets that are (a) large, because they affect the apparent luminosity of the star more and (b) close to their sun because they transit more frequently.

What is meant is that assuming 2 planets of the same diameter, one being closer to the star than the other, the closer one blots out more of the star's visible "disk".

Its orbit being closer in will likely be much smaller and will indicate a much faster revolution around the star it orbits. Its transit across the disk of the star will be much quicker than the same sized plant further out from the star.

Distance from the star does not change the amount of light that the planet blocks from our perspective. The other effect matters. Kepler is biased to find planets that are (a) large, because they affect the apparent luminosity of the star more and (b) close to their sun because they transit more frequently.

I agree with your points (a) and (b) but not with: "Distance from the star does not change the amount of light that the planet blocks from our perspective."

Despite the greater distances involved, Kepler is designed to detect infinitesimal variations in light output from a star. If two Earth-sized planets were orbiting the same star, one at Earth's orbital distance, and one at Neptune's orbital distance, how could anyone claim that the one at Neptune's distance wouldn't appear to block more light when it passed into the field of view between the star and Kepler?

If Kepler were much closer to the star system, there's no question this would be true (much like my Solar Eclipse vs. Vanus transit analogy.) Is there something that happens at great distances that makes this not true, because so far no one has been able to provide definitive proof of this.

I agree, at the distances involved, the change would be minimal, but that's exactly what Kepler is designed to detect. Also, I should note that in the case of the Neptune-orbital-distance planet, the time between detection would be much longer, slowing Kepler's ability to confirm its existence, but that's not a point of contention here.

I agree, at the distances involved, the change would be minimal, but that's exactly what Kepler is designed to detect.

"minimal" doesn't even begin to describe how small the difference would be - don't forget that it's not just the measuring accuracy of Kepler that plays a role here - consider sunspots. Even the smallest sunspot ever observed on our own sun would influence the observed occultation brightness in this case thousands, if not hundreds of thousands of times more than the distance-difference effect would.

In essense, measuring the difference between identically sized planets 10 & 100 million km away from Kepler 37a by the "size" of the eclipse is impossible because the random brightness fluctuations of the star itself will be so much larger than the difference that Kepler will not be able to detect the effect even if it theoretically had the required sensitivity to do so.

And just to be clear, Kepler can *barely* detect this tiny planet crossing the star's disk, and it can only detect it by combining measurements over many, many crossings - understand this : Kepler cannot detect a single crossing of this tiny planet - it needs to average out measurements over multiple crossings to achieve the necessary detection threshold, and this threshold is a signal millions of times larger than the difference that would occur between planets 10 & 100 million km away from the star.

Bottom line: the sentence in the original article is *wrong* - if Kepler were capable of detecting such small differences, it would be capable of detecting asteroids the size of the Chelyabinsk meteor around other stars - it doesn't and couldn't, even with a *perfect* detector, because the star itself is also a source of noise.

I agree, at the distances involved, the change would be minimal, but that's exactly what Kepler is designed to detect. Also, I should note that in the case of the Neptune-orbital-distance planet, the time between detection would be much longer, slowing Kepler's ability to confirm its existence, but that's not a point of contention here.

No, they're minimal even in comparison to the changes Kepler is designed to detect. Take a Sol-like star 10 light years away, with two identical Earth-like planets, one at 1 AU and 1 at 100 AU. The difference in brightness during transit would be around 0.03 parts per million. The uncertainty in the radius of, for instance, KOI-172.02 is about ten million times that.

I agree, at the distances involved, the change would be minimal, but that's exactly what Kepler is designed to detect. Also, I should note that in the case of the Neptune-orbital-distance planet, the time between detection would be much longer, slowing Kepler's ability to confirm its existence, but that's not a point of contention here.

No, they're minimal even in comparison to the changes Kepler is designed to detect. Take a Sol-like star 10 light years away, with two identical Earth-like planets, one at 1 AU and 1 at 100 AU. The difference in brightness during transit would be around 0.03 parts per million. The uncertainty in the radius of, for instance, KOI-172.02 is about ten million times that.

Actual facts, that's what I like to see. Going further with your point, how, then, could the author claim that objects with a closer orbit result in a larger eclipse, making them more detectable by Kepler?

The only solution I can see is that the author misappropriated the reasoning for why planets with closer orbital distances are more detectable by Kepler, i.e. more frequent passes in front of the host star, allowing for faster verification of existence, vs. the incorrect attribution of "larger eclipse at closer distances."

I agree, at the distances involved, the change would be minimal, but that's exactly what Kepler is designed to detect.

"minimal" doesn't even begin to describe how small the difference would be - don't forget that it's not just the measuring accuracy of Kepler that plays a role here - consider sunspots. Even the smallest sunspot ever observed on our own sun would influence the observed occultation brightness in this case thousands, if not hundreds of thousands of times more than the distance-difference effect would.

In essense, measuring the difference between identically sized planets 10 & 100 million km away from Kepler 37a by the "size" of the eclipse is impossible because the random brightness fluctuations of the star itself will be so much larger than the difference that Kepler will not be able to detect the effect even if it theoretically had the required sensitivity to do so.

And just to be clear, Kepler can *barely* detect this tiny planet crossing the star's disk, and it can only detect it by combining measurements over many, many crossings - understand this : Kepler cannot detect a single crossing of this tiny planet - it needs to average out measurements over multiple crossings to achieve the necessary detection threshold, and this threshold is a signal millions of times larger than the difference that would occur between planets 10 & 100 million km away from the star.

Bottom line: the sentence in the original article is *wrong* - if Kepler were capable of detecting such small differences, it would be capable of detecting asteroids the size of the Chelyabinsk meteor around other stars - it doesn't and couldn't, even with a *perfect* detector, because the star itself is also a source of noise.

My original assumption was basically this. It just seems like common sense to me and I struggled to understand the author's statement. However, since I'm just an armchair astrophysicist I figured it was me that was wrong. However yours and Christopher's arguments are very persuasive.

I'd really like to hear from the Ars science team or the author, and am a little shocked we haven't already. Was the statement a mistake? If not, can you try to explain it a little more clearly?

One possible confusion: at the other side of the orbit, with the planet passing behind the star, there is a signal that does increase in intensity with proximity to the star due to reflected light from the planet. Similarly, the changing phase of the planet alone gives a signal that is stronger for closer (and thus more brightly illuminated) planets and potentially detectable even when the planet isn't aligned well enough to actually transit the star. I don't know if Kepler is capable of spotting this signal in even extreme cases, though. It is designed specifically to look for the much stronger signal a transit causes.

One possible confusion: at the other side of the orbit, with the planet passing behind the star, there is a signal that does increase in intensity with proximity to the star due to reflected light from the planet. Similarly, the changing phase of the planet alone gives a signal that is stronger for closer (and thus more brightly illuminated) planets and potentially detectable even when the planet isn't aligned well enough to actually transit the star. I don't know if Kepler is capable of spotting this signal in even extreme cases, though. It is designed specifically to look for the much stronger signal a transit causes.

Kepler is designed to detect brightness variations regardless of cause (would be difficult to do otherwise anyway) - so yes, changing phases and illuminated when they are moving behind their star would be detectable - in fact this effect has been seen for a long time in cases of close binary stars where one of the stars is much brighter than the other.

However, the same argument applies : we are talking about variations orders of magnitude smaller than the eclipses themselves, so you'd only be able to detect them on very big planets very close to their parent star, and only planets which reflect a lot of the infalling light (our moon, for example, absorbs 90% of infalling light and is actually very dark - the "albedo" of the moon is 0.11).

So the size of this effect is not just influenced by the planet's distance from the star and its diameter but also by its albedo value.